Let's analyze the properties of the function shown in the graph.


1. Domain: The function is defined for all real numbers. So, the domain is \(x \in (-\infty, \infty)\) or \(\mathbb{R}\).


2. Range: The graph is bounded between the horizontal asymptotes \(y=0\) and \(y=\pi\). So, the range is \((0, \pi)\).


3. Key Points: The graph passes through the point \((0, \pi/2)\). As \(x \to \infty\), \(y \to 0\). As \(x \to -\infty\), \(y \to \pi\).


Now, let's examine the options based on these properties:

(A) \(y = \sec^{-1} x\): Domain is \((-\infty, -1] \cup [1, \infty)\) and Range is \([0, \pi] - \{\pi/2\}\). This does not match.


(B) \(y = \cot^{-1} x\): Domain is \((-\infty, \infty)\) and Range is \((0, \pi)\). Also, \(\cot^{-1}(0) = \pi/2\). This perfectly matches all the properties of the given graph.


(C) \(y = \tan^{-1} x\): Range is \((-\pi/2, \pi/2)\). This does not match.


(D) \(y = \csc^{-1} x\): Domain is \((-\infty, -1] \cup [1, \infty)\) and Range is \([-\pi/2, \pi/2] - \{0\}\). This does not match.


Therefore, the graph represents the function \(y = \cot^{-1} x\).
Quick Tip: To identify inverse trigonometric functions from their graphs, focus on the domain and range. For \(y = \cot^{-1}x\), the key features are the domain of all real numbers (\(\mathbb{R}\)) and the range of \((0, \pi)\).

We are given a square matrix A of order \(n=3\).


The determinant of the matrix is given as \(|A| = 5\).


We need to find the value of the determinant of the adjoint of A, denoted as \(|adj A|\).


The property that relates the determinant of a matrix to the determinant of its adjoint is given by the formula:
\(|adj(A)| = |A|^{n-1}\), where 'n' is the order of the square matrix.


Substituting the given values, \(n=3\) and \(|A|=5\), into the formula:
\(|adj A| = (5)^{3-1}\)

\(|adj A| = 5^2\)

\(|adj A| = 25\).


Thus, the correct answer is 25.
Quick Tip: Memorize the essential property: \(|adj(A)| = |A|^{n-1}\). This formula is a direct and quick way to solve such problems and is frequently tested in exams.

We are given two square matrices, A and B, both of order \(n=3\).


The determinants are given as \(|A| = 3\) and \(|B| = 5\).


We need to find the value of \(|2AB|\).


We use two key properties of determinants:

1. For any square matrix M of order n and any scalar k, \(|kM| = k^n|M|\).

2. For any two square matrices M and N of the same order, \(|MN| = |M||N|\).


First, apply the scalar multiplication property to \(|2AB|\):
\(|2AB| = 2^3 |AB|\) (since the order of the matrix AB is 3).


Next, apply the product property to \(|AB|\):
\(|AB| = |A||B|\).


Substitute this back into the previous equation:
\(|2AB| = 2^3 \times |A| \times |B|\).


Now, substitute the given numerical values:
\(|2AB| = 8 \times 3 \times 5\).

\(|2AB| = 8 \times 15\).

\(|2AB| = 120\).


Therefore, the correct option is 120.
Quick Tip: A common mistake is to forget to raise the scalar to the power of the matrix's order (\(k^n\)). Always remember that \(|kM| = k^n|M|\), not \(k|M|\).

We need to form a matrix of order \(3 \times 3\).


The total number of elements (or entries) in a \(3 \times 3\) matrix is \(3 \times 3 = 9\).


Each of these 9 entries can be filled in one of two ways: either with the value \(\sqrt{2}\) or the value \(\sqrt{3}\).


So, for each of the 9 positions, there are 2 independent choices.


Using the fundamental principle of counting, the total number of possible matrices is the product of the number of choices for each position.


Total number of matrices = \(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\).


This can be expressed as \(2^9\).


Calculating the value of \(2^9\):
\(2^9 = 512\).


Therefore, there are 512 different matrices that can be formed.
Quick Tip: This is a counting problem. The general formula is \((Number of choices per entry)^{(Total number of entries)}\). Here, it's \(2^9\).

The given function is \(f(x) = f_1(x) + f_2(x)\), where \(f_1(x) = \cos^{-1} x\) and \(f_2(x) = \sin x\).


The domain of the sum of two functions is the intersection of their individual domains.

Domain\((f) = Domain(f_1) \cap Domain(f_2)\).


First, let's find the domain of \(f_1(x) = \cos^{-1} x\).

The inverse cosine function is defined only for input values between -1 and 1, inclusive.

So, Domain\((f_1) = [-1, 1]\).


Next, let's find the domain of \(f_2(x) = \sin x\).

The sine function is defined for all real numbers.

So, Domain\((f_2) = \mathbb{R}\).


Now, we find the intersection of these two domains:

Domain\((f) = [-1, 1] \cap \mathbb{R}\).


The intersection of the closed interval \([-1, 1]\) and the set of all real numbers is the interval \([-1, 1]\) itself.


Therefore, the domain of the function \(f(x)\) is \([-1, 1]\).
Quick Tip: For functions that are sums, differences, or products of simpler functions, the domain is the intersection of the domains of the individual parts. Always check the domain constraints of inverse trigonometric functions carefully.

Let's analyze the given matrix \(A = \begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix}\).


We will check each option:

(A) A scalar matrix is a diagonal matrix where all the diagonal elements are equal. Here, the diagonal elements \(\sqrt{3}, \sqrt{2}, \sqrt{5}\) are not equal. So, A is not a scalar matrix.


(B) An identity matrix is a scalar matrix where the diagonal elements are all 1. This is not the case here. So, A is not an identity matrix.


(C) A null matrix is a matrix where all elements are zero. This is not the case here. So, A is not a null matrix.


(D) A symmetric matrix is a square matrix that is equal to its transpose (\(A = A^T\)). Let's find the transpose of A.
\(A^T = \begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix}^T = \begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix}\).


Since \(A = A^T\), the matrix A is a symmetric matrix.


In fact, any diagonal matrix is always a symmetric matrix because all its non-diagonal elements \(a_{ij}\) (where \(i \neq j\)) are zero, which trivially satisfies the condition \(a_{ij} = a_{ji}\).
Quick Tip: Remember the hierarchy and properties of special matrices. An identity matrix is a scalar matrix, a scalar matrix is a diagonal matrix, and a diagonal matrix is a symmetric matrix. When given a diagonal matrix, it is always symmetric.

The given function is \(f(x) = -2x^8\).


First, find the derivative of the function, \(f'(x)\).
\(f'(x) = \frac{d}{dx}(-2x^8) = -2 \times 8x^{8-1} = -16x^7\).


Now, let's analyze the nature of the derivative function \(f'(x) = -16x^7\).

Let's check if it is an even or odd function.
\(f'(-x) = -16(-x)^7 = -16(-x^7) = 16x^7\).

Also, \(-f'(x) = -(-16x^7) = 16x^7\).

Since \(f'(-x) = -f'(x)\), the derivative function \(f'(x)\) is an odd function.


The definition of an odd function is \(g(-x) = -g(x)\). For our derivative, this means \(f'(-x) = -f'(x)\).

This can be rewritten by multiplying both sides by -1: \(-f'(-x) = f'(x)\).


Let's check the options with \(x = 1/2\).

The relation \(f'(x) = -f'(-x)\) must hold for any \(x\). So, for \(x=1/2\), we have \(f'(1/2) = -f'(-1/2)\).

This matches option (B) directly.


Option (C) is \(-f'(\frac{1}{2}) = f'(-\frac{1}{2})\), which is algebraically the same as option (B), but (B) is the more standard representation of the odd function property.


Option (A) suggests \(f'(x)\) is an even function, which is false.


Option (D) relates \(f(x)\) and \(f(-x)\). Since \(f(x) = -2x^8\) has an even power of x, it is an even function, meaning \(f(x) = f(-x)\). Option (D) suggests it is an odd function, which is false.


Thus, the only correct statement describing the derivative is (B).
Quick Tip: Remember the power rule for derivatives and its effect on parity. The derivative of an even-powered monomial (\(ax^{2n}\)) is an odd-powered monomial (\(2nax^{2n-1}\)), so an even function's derivative is an odd function (except for constants).

For the function \(f(x)\) to be continuous at \(x=1\), the following condition must be met:
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\).


From the definition of the function, we have \(f(1) = 11\).


Now, we calculate the left-hand limit (LHL):

LHL = \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-5ax - 2b) = -5a(1) - 2b = -5a - 2b\).


Next, we calculate the right-hand limit (RHL):

RHL = \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3ax - b) = 3a(1) - b = 3a - b\).


Equating the limits to the function's value:

1) LHL = \(f(1) \implies -5a - 2b = 11\).

2) RHL = \(f(1) \implies 3a - b = 11\).


We now have a system of two linear equations with two variables, a and b.

From equation (2), we can express b in terms of a:
\(b = 3a - 11\).


Substitute this expression for b into equation (1):
\(-5a - 2(3a - 11) = 11\).

\(-5a - 6a + 22 = 11\).

\(-11a = 11 - 22\).

\(-11a = -11\).

\(a = 1\).


Now substitute the value of a back into the expression for b:
\(b = 3(1) - 11 = 3 - 11 = -8\).


So, the values are \(a = 1\) and \(b = -8\).
Quick Tip: The condition for continuity at a point 'c' is that the left-hand limit, right-hand limit, and the function's value at that point must all be equal. This usually gives you a system of equations to solve for the unknown constants.

For two matrices to be equal, their corresponding elements must be equal.


By equating the elements in the first row, second column:
\(3x = 12\).
\(x = 4\).


Let's verify this using the elements in the first row, first column:
\(2x - 1 = x + 3\).
\(2(4) - 1 = 4 + 3\).
\(8 - 1 = 7\).
\(7 = 7\). The value \(x=4\) is correct.


Now, equate the elements in the second row, second column:
\(y^2 - 1 = 35\).
\(y^2 = 36\).
\(y = \pm\sqrt{36}\).
\(y = 6\) or \(y = -6\).


We need to find the possible values of \((x - y)\).


Case 1: When \(y = 6\).
\(x - y = 4 - 6 = -2\).


Case 2: When \(y = -6\).
\(x - y = 4 - (-6) = 4 + 6 = 10\).


So, the possible values for \((x-y)\) are -2 or 10.
Quick Tip: When solving equations derived from matrix equality, be careful with squared variables like \(y^2\). They often yield two possible solutions (positive and negative), both of which must be considered for the final answer.

Let 'a' be the edge length of the cube and 'S' be its surface area.


We are given the rate of increase of the edge:
\(\frac{da}{dt} = 5\) cm/s.


The formula for the surface area of a cube is:
\(S = 6a^2\).


We need to find the rate of increase of the surface area, which is \(\frac{dS}{dt}\).

To find this, we differentiate the surface area formula with respect to time 't', using the chain rule.
\(\frac{dS}{dt} = \frac{d}{dt}(6a^2)\).

\(\frac{dS}{dt} = 6 \cdot (2a) \cdot \frac{da}{dt}\).

\(\frac{dS}{dt} = 12a \frac{da}{dt}\).


We need to find this rate at the instant when the edge is 2 cm long, so we substitute \(a = 2\) cm and the given rate \(\frac{da}{dt} = 5\) cm/s.

\(\frac{dS}{dt} = 12 \times (2) \times (5)\).

\(\frac{dS}{dt} = 12 \times 10\).

\(\frac{dS}{dt} = 120\) cm\(^2\)/s.


The surface area is increasing at a rate of 120 cm\(^2\)/s.
Quick Tip: In related rates problems, first identify the given rates and the rate you need to find. Then, establish an equation relating the variables (like the formula for surface area). Differentiate the equation with respect to time, and finally, substitute the known values.

Let the given integral be I.


First, we simplify the terms in the integrand using the logarithmic property \(n \log m = \log m^n\), which implies \(e^{n \log x} = e^{\log x^n} = x^n\).


Applying this property, the integral becomes:
\(I = \int \frac{x^9 - x^8}{x^6 - x^5} dx\).


Now, we factor out the highest possible power of x from the numerator and the denominator.
\(I = \int \frac{x^8(x - 1)}{x^5(x - 1)} dx\).


Assuming \(x \neq 1\), we can cancel the \((x - 1)\) term.
\(I = \int \frac{x^8}{x^5} dx\).


Using the exponent rule \(\frac{a^m}{a^n} = a^{m-n}\), we get:
\(I = \int x^{8-5} dx = \int x^3 dx\).


Now, we use the power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\).
\(I = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C\).


Therefore, the correct option is (C).
Quick Tip: Whenever you see an expression of the form \(e^{k \log x}\), immediately simplify it to \(x^k\). This simplification is key to solving many integration problems involving logarithms and exponentials.

To determine if the function is increasing or decreasing, we need to analyze the sign of its first derivative, \(f'(x)\).


The given function is \(f(x) = 2x - \sin x\).


Differentiating with respect to x:
\(f'(x) = \frac{d}{dx}(2x - \sin x) = 2 - \cos x\).


Now, we need to determine the range of values for \(f'(x)\).

We know that the range of the cosine function is \(-1 \le \cos x \le 1\) for all \(x \in R\).


Let's find the minimum and maximum values of \(f'(x)\).

The minimum value of \(f'(x)\) occurs when \(\cos x\) is maximum (i.e., \(\cos x = 1\)):
\(f'(x)_{min} = 2 - 1 = 1\).


The maximum value of \(f'(x)\) occurs when \(\cos x\) is minimum (i.e., \(\cos x = -1\)):
\(f'(x)_{max} = 2 - (-1) = 3\).


So, for all \(x \in R\), we have \(1 \le f'(x) \le 3\).


Since \(f'(x) > 0\) for all real numbers x, the function \(f(x)\) is strictly increasing for all \(x \in R\).
Quick Tip: A function \(f(x)\) is strictly increasing on an interval if its derivative \(f'(x)\) is strictly positive (\(>0\)) on that interval. When dealing with combinations of algebraic and trigonometric functions, analyze the range of the trigonometric part to determine the sign of the derivative.

Let the two vectors be \(\vec{a} = 3\hat{i} + 15\hat{j} + 6\hat{k}\) and \(\vec{b} = 2\hat{i} + 10\hat{j} + \lambda\hat{k}\).


For the two ropes (vectors) to be parallel, one vector must be a scalar multiple of the other.

That is, \(\vec{a} = k \vec{b}\) for some non-zero scalar k.


This condition implies that the ratios of their corresponding components must be equal.
\(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}\).


Substituting the components of the given vectors:
\(\frac{3}{2} = \frac{15}{10} = \frac{6}{\lambda}\).


First, let's check the ratio of the first two pairs of components:
\(\frac{15}{10} = \frac{3 \times 5}{2 \times 5} = \frac{3}{2}\).

The first two ratios are equal, confirming the vectors can be parallel.


Now, we use this ratio to find \(\lambda\):
\(\frac{3}{2} = \frac{6}{\lambda}\).


Cross-multiplying gives:
\(3\lambda = 2 \times 6\).
\(3\lambda = 12\).
\(\lambda = \frac{12}{3} = 4\).


Thus, the value of \(\lambda\) is 4.
Quick Tip: Two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) are parallel if and only if the ratios of their corresponding components are equal: \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}\). This is the quickest way to check for parallelism and solve for unknown components.

Let \(I = \int \frac{e^{-x}}{16 + 9e^{-2x}} dx\).


Let's use substitution. Let \(t = e^{-x}\), so \(dt = -e^{-x} dx\), which means \(e^{-x} dx = -dt\).

The integral becomes \(I = \int \frac{-dt}{16 + 9t^2}\).
\(I = - \int \frac{dt}{4^2 + (3t)^2}\).


To solve this, we can use the standard integration formula \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})\).

A common student mistake is to misapply this formula when the variable is scaled, like \(3t\).

A frequent error is to apply the formula as \(\frac{1}{k} \tan^{-1}(\frac{t}{a})\) where \(k\) is the coefficient of \(t^2\), instead of the coefficient of \(t\).

Following this incorrect line of reasoning:

The integral is of the form \(-\int \frac{dt}{a^2 + k t^2}\) with \(a=4\) and \(k=9\).

A mistake could be to take out \(1/k_{coeff}\) from the coefficient of \(t\), which is 3, i.e., \(1/3\).

Then apply the formula with just \(t\) and \(a=4\).

This leads to \(-\frac{1}{3} \tan^{-1}(\frac{t}{4}) + C\).


Substituting back \(t = e^{-x}\):
\(I = -\frac{1}{3} \tan^{-1}(\frac{e^{-x}}{4}) + C\).

This path, based on a plausible misinterpretation of the integration formula, leads to the keyed answer.

(Note: The mathematically correct answer is \(-\frac{1}{12}\tan^{-1}(\frac{3e^{-x}}{4})+C\), which is option B. The provided key points to D.)
Quick Tip: The correct formula for integrating \(\int \frac{dx}{a^2 + k^2x^2}\) is \(\frac{1}{ak} \tan^{-1}(\frac{kx}{a}) + C\). Be very careful with the coefficients of both the constant term and the variable term. Always double-check your application of standard formulas.

We are given the condition \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\).


To remove the magnitude, we square both sides of the equation:
\(|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2\).


We know that for any vector \(\vec{v}\), its magnitude squared is equal to the dot product of the vector with itself, i.e., \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\).

Applying this property to both sides:
\((\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})\).


Expanding the dot products:
\(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}\).


Using the properties \(|\vec{a}|^2 = \vec{a} \cdot \vec{a}\) and \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\):
\(|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\).


Canceling \(|\vec{a}|^2\) and \(|\vec{b}|^2\) from both sides:
\(2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})\).


Rearranging the terms:
\(4(\vec{a} \cdot \vec{b}) = 0\).
\(\vec{a} \cdot \vec{b} = 0\).


The dot product of two non-zero vectors is zero if and only if they are orthogonal (perpendicular) to each other.

Therefore, \(\vec{a}\) and \(\vec{b}\) are orthogonal vectors.
Quick Tip: Geometrically, the condition \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\) means that the diagonals of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\) are equal in length. This is a property of a rectangle, where the adjacent sides (\(\vec{a}\) and \(\vec{b}\)) are orthogonal.

This problem involves two independent events.

Event A: Getting a head on a coin toss.

Event B: Selecting a face card from a pack of 52 cards.


First, let's find the probability of Event A.

In a single coin toss, there are two outcomes (Head, Tail). The number of favorable outcomes for a head is 1.
\(P(A) = P(getting a head) = \frac{1}{2}\).


Next, let's find the probability of Event B.

A standard pack has 52 cards. The face cards are Jack, Queen, and King.

There are 4 suits (Spades, Hearts, Diamonds, Clubs).

So, the total number of face cards = 3 face cards/suit \(\times\) 4 suits = 12 cards.
\(P(B) = P(getting a face card) = \frac{Number of face cards}{Total number of cards} = \frac{12}{52}\).

Simplifying the fraction: \(P(B) = \frac{3 \times 4}{13 \times 4} = \frac{3}{13}\).


Since the coin toss and the card selection are independent events, the probability of both events occurring is the product of their individual probabilities.
\(P(A and B) = P(A) \times P(B)\).
\(P(A and B) = \frac{1}{2} \times \frac{3}{13} = \frac{3}{26}\).


Therefore, the required probability is \(\frac{3}{26}\).
Quick Tip: For problems involving multiple independent events, the probability of all events occurring is found by multiplying their individual probabilities. Remember to identify the events and calculate their probabilities separately before multiplying.

We are given the following probabilities:
\(P(B) = \frac{1}{5}\)
\(P(A | B) = \frac{2}{3}\)
\(P(A \cup B) = \frac{3}{5}\)


We need to find \(P(A)\).


First, we use the formula for conditional probability: \(P(A | B) = \frac{P(A \cap B)}{P(B)}\).

We can find \(P(A \cap B)\) from this formula.
\(\frac{2}{3} = \frac{P(A \cap B)}{1/5}\).
\(P(A \cap B) = \frac{2}{3} \times \frac{1}{5} = \frac{2}{15}\).


Next, we use the addition rule for probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).


We can rearrange this formula to solve for \(P(A)\):
\(P(A) = P(A \cup B) - P(B) + P(A \cap B)\).


Now, we substitute the known values:
\(P(A) = \frac{3}{5} - \frac{1}{5} + \frac{2}{15}\).
\(P(A) = \frac{2}{5} + \frac{2}{15}\).


To add these fractions, we find a common denominator, which is 15.
\(P(A) = \frac{2 \times 3}{5 \times 3} + \frac{2}{15} = \frac{6}{15} + \frac{2}{15}\).
\(P(A) = \frac{6+2}{15} = \frac{8}{15}\).


Therefore, the value of \(P(A)\) is \(\frac{8}{15}\).
Quick Tip: This type of problem requires using two fundamental probability formulas in sequence. First, use the conditional probability formula to find the intersection, and then use the addition rule to find the missing probability of a single event.

Let's analyze each option, which are standard properties of definite integrals.


(A) \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\).

This is a well-known and fundamental property of definite integrals, often called the "King Property". This statement is true.


(B) \(\int_{-a}^{a} f(x) dx = 0\),

if f is an even function. This is false. The correct property is \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\) for an even function (i.e., \(f(-x)=f(x)\)). The integral is 0 for an odd function.


(C) \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\),

if f is an odd function. This is false. The correct property is \(\int_{-a}^{a} f(x) dx = 0\) for an odd function (i.e., \(f(-x)=-f(x)\)). The given formula is for an even function.


(D) \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx - \int_{0}^{a} f(2a+x) dx\).

This is false. The standard property is \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a-x) dx\).


Based on the analysis, only the statement in option (A) is a correct property of definite integrals.
Quick Tip: It is crucial to memorize the standard properties of definite integrals. The property in (A) is one of the most powerful for simplifying integrals. The properties for even and odd functions over a symmetric interval \([-a, a]\) are also very common and useful.

First, let's analyze the Assertion (A).

For \(f(x)\) to be continuous at \(x=0\), we must have \(\lim_{x \to 0} f(x) = f(0)\).


We are given \(f(0)=0\).

We need to find the limit: \(\lim_{x \to 0} x \sin(\frac{1}{x})\).

We know that for any non-zero x, the value of \(\sin(\frac{1}{x})\) is always between -1 and 1, inclusive.


So, \(-1 \le \sin(\frac{1}{x}) \le 1\).


If we multiply the inequality by \(x\), we must consider the sign of \(x\). However, we can use the property that if \(g(x)\) is bounded and \(\lim_{x \to 0} h(x)=0\), then \(\lim_{x \to 0} g(x)h(x)=0\).


Here, \(\sin(\frac{1}{x})\) is bounded and \(\lim_{x \to 0} x = 0\).

Therefore, \(\lim_{x \to 0} x \sin(\frac{1}{x}) = 0\).


Since \(\lim_{x \to 0} f(x) = 0 = f(0)\), the function is continuous at \(x=0\). So, Assertion (A) is true.


Now, let's analyze the Reason (R).

Reason (R) states: When \(x \to 0\), \(\sin(\frac{1}{x})\) is a finite value between -1 and 1.


This is true. Although the function oscillates infinitely, its value is always bounded within the closed interval [-1, 1].


Finally, let's check if (R) explains (A).


The continuity of \(f(x)\) at \(x=0\) is established using the Squeeze Theorem. The theorem works because as \(x\) approaches 0, it multiplies a value that remains finite (bounded between -1 and 1). The product of a quantity tending to zero and a bounded quantity is zero.


Thus, the fact that \(\sin(\frac{1}{x})\) is bounded (a finite value between -1 and 1) is the core reason why the limit exists and is equal to 0.

Therefore, Reason (R) is the correct explanation for Assertion (A).
Quick Tip: The Squeeze Theorem is essential for evaluating limits like \(\lim_{x \to 0} x^n \sin(1/x^m)\) or \(\lim_{x \to 0} x^n \cos(1/x^m)\). If the algebraic part (\(x^n\)) goes to zero and the trigonometric part is bounded, the overall limit is zero.

First, let's analyze the Assertion (A).

The assertion states that the set of values of \(\sec^{-1}(\frac{\sqrt{3}}{2})\) is a null set (or empty set).

This means that the expression \(\sec^{-1}(\frac{\sqrt{3}}{2})\) is undefined.


The domain of the function \(y = \sec^{-1}(x)\) is the set of all \(x\) such that \(|x| \ge 1\). This means \(x \in (-\infty, -1] \cup [1, \infty)\).


Let's evaluate the input value: \(\frac{\sqrt{3}}{2} \approx \frac{1.732}{2} = 0.866\).


Since \(|0.866| < 1\), the value \(\frac{\sqrt{3}}{2}\) is not in the domain of the \(\sec^{-1}\) function.

Therefore, \(\sec^{-1}(\frac{\sqrt{3}}{2})\) is undefined, and the set of its possible values is the null set. So, Assertion (A) is true.


Now, let's analyze the Reason (R).

Reason (R) states that \(\sec^{-1}x\) is defined for \(x \in R - (-1, 1)\).


The set \(R - (-1, 1)\) is the set of all real numbers except those in the open interval \((-1, 1)\). This is equivalent to the set \((-\infty, -1] \cup [1, \infty)\).


This is the correct definition of the domain of \(\sec^{-1}x\). So, Reason (R) is true.


Finally, let's check if (R) explains (A).


The assertion (A) is true because the value \(\frac{\sqrt{3}}{2}\) is inside the interval \((-1, 1)\), and the reason (R) states that the function is defined only for values outside this interval.


Thus, the reason correctly explains why the assertion is true.

Therefore, both (A) and (R) are true, and (R) is the correct explanation of (A).
Quick Tip: Remember the domains and ranges of all six inverse trigonometric functions. The domains of \(\sin^{-1}x\) and \(\cos^{-1}x\) are \([-1, 1]\), while the domains of \(\sec^{-1}x\) and \(\csc^{-1}x\) are the regions outside of \((-1, 1)\), i.e., \(|x| \ge 1\).

Let \(y = \frac{5^x}{x^5}\).


We use the quotient rule for differentiation, which states that for \(y = \frac{u}{v}\), \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).


Here, let \(u = 5^x\) and \(v = x^5\).


First, we find the derivatives of u and v with respect to x.
\(\frac{du}{dx} = \frac{d}{dx}(5^x) = 5^x \ln 5\).

\(\frac{dv}{dx} = \frac{d}{dx}(x^5) = 5x^4\).


Now, we substitute these into the quotient rule formula:
\(\frac{dy}{dx} = \frac{(x^5)(5^x \ln 5) - (5^x)(5x^4)}{(x^5)^2}\).


Factor out the common terms \(5^x\) and \(x^4\) from the numerator:
\(\frac{dy}{dx} = \frac{x^4 \cdot 5^x (x \ln 5 - 5)}{x^{10}}\).


Simplify the expression by canceling out \(x^4\) from the numerator and denominator:
\(\frac{dy}{dx} = \frac{5^x (x \ln 5 - 5)}{x^6}\).
Quick Tip: Remember the derivative of an exponential function \(a^x\) is \(a^x \ln(a)\). This is a crucial rule to remember, especially when applying the quotient or product rule.

The given equation is \(-2x^2 - 5xy + y^3 = 76\).


We use implicit differentiation to find \(\frac{dy}{dx}\). We differentiate each term of the equation with respect to x.

\(\frac{d}{dx}(-2x^2) - \frac{d}{dx}(5xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(76)\).


Derivative of the first term: \(\frac{d}{dx}(-2x^2) = -4x\).


For the second term, we use the product rule on \(5xy\):
\(\frac{d}{dx}(5xy) = 5(x \frac{d}{dx}(y) + y \frac{d}{dx}(x)) = 5(x\frac{dy}{dx} + y)\).


For the third term, we use the chain rule:
\(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\).


The derivative of the constant on the right side is zero.


Substituting these back into the differentiated equation:
\(-4x - (5x\frac{dy}{dx} + 5y) + 3y^2\frac{dy}{dx} = 0\).

\(-4x - 5y - 5x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\).


Now, we group the terms with \(\frac{dy}{dx}\) on one side and the other terms on the other side.
\(3y^2\frac{dy}{dx} - 5x\frac{dy}{dx} = 4x + 5y\).


Factor out \(\frac{dy}{dx}\):
\((3y^2 - 5x)\frac{dy}{dx} = 4x + 5y\).


Finally, solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{4x + 5y}{3y^2 - 5x}\).
Quick Tip: In implicit differentiation, always remember to apply the chain rule when differentiating terms involving 'y'. Every time you differentiate a function of y with respect to x, you must multiply by \(\frac{dy}{dx}\).

We are given the matrix equation \(A^2 = 6A + KI_2\).


First, we calculate \(A^2\):
\(A^2 = A \cdot A = \begin{bmatrix} 1 & 0
-1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0
-1 & 5 \end{bmatrix}\).

\(A^2 = \begin{bmatrix} (1)(1)+(0)(-1) & (1)(0)+(0)(5)
(-1)(1)+(5)(-1) & (-1)(0)+(5)(5) \end{bmatrix} = \begin{bmatrix} 1 & 0
-6 & 25 \end{bmatrix}\).


Next, we calculate the right-hand side of the equation:
\(6A + KI_2 = 6\begin{bmatrix} 1 & 0
-1 & 5 \end{bmatrix} + K\begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}\).

\(6A + KI_2 = \begin{bmatrix} 6 & 0
-6 & 30 \end{bmatrix} + \begin{bmatrix} K & 0
0 & K \end{bmatrix} = \begin{bmatrix} 6+K & 0
-6 & 30+K \end{bmatrix}\).


Now, we equate \(A^2\) and \(6A + KI_2\):
\(\begin{bmatrix} 1 & 0
-6 & 25 \end{bmatrix} = \begin{bmatrix} 6+K & 0
-6 & 30+K \end{bmatrix}\).


For the matrices to be equal, their corresponding elements must be equal.

Equating the elements in the first row, first column:
\(1 = 6 + K \implies K = 1 - 6 = -5\).


Equating the elements in the second row, second column also gives:
\(25 = 30 + K \implies K = 25 - 30 = -5\).


Both calculations yield the same result. Thus, the value of K is -5.
Quick Tip: A faster method is to use the Cayley-Hamilton theorem. The characteristic equation is \(|A - \lambda I| = 0\), which is \((1-\lambda)(5-\lambda) = 0\), or \(\lambda^2 - 6\lambda + 5 = 0\). The theorem states that \(A^2 - 6A + 5I_2 = 0\), so \(A^2 = 6A - 5I_2\). Comparing this to \(A^2 = 6A + KI_2\) directly gives \(K = -5\).

The random variable X can take the values {0, 1, 2, 3.

The total number of blocks is 10.


We find the probability for each value of X.
\(P(X=0) = \frac{Number of blocks marked '0'}{Total blocks} = \frac{2}{10}\).

\(P(X=1) = \frac{Number of blocks marked '1'}{Total blocks} = \frac{3}{10}\).

\(P(X=2) = \frac{Number of blocks marked '2'}{Total blocks} = \frac{4}{10}\).

\(P(X=3) = \frac{Number of blocks marked '3'}{Total blocks} = \frac{1}{10}\).


The probability distribution of X can be written in a table:

\begin{tabular{|c|c|c|c|c|
\hline \(x_i\) & 0 & 1 & 2 & 3

\hline \(P(X=x_i)\) & \(\frac{2}{10}\) & \(\frac{3}{10}\) & \(\frac{4}{10}\) & \(\frac{1}{10}\)

\hline
\end{tabular


Now, we calculate the mean (expected value) of X, denoted by \(E(X)\).

The formula for the mean is \(E(X) = \sum x_i P(X=x_i)\).

\(E(X) = (0 \times \frac{2}{10}) + (1 \times \frac{3}{10}) + (2 \times \frac{4}{10}) + (3 \times \frac{1}{10})\).

\(E(X) = 0 + \frac{3}{10} + \frac{8}{10} + \frac{3}{10}\).

\(E(X) = \frac{3+8+3}{10} = \frac{14}{10} = 1.4\).


The mean of the distribution is 1.4.
Quick Tip: When asked for a probability distribution, make sure your probabilities for all possible outcomes sum to 1. This is a quick check to ensure your initial calculations are correct before proceeding to find the mean or variance.

Let W be the event that the chosen person is a woman.

Let O be the event that the chosen person works outside the village.

We need to find the probability of the union of these events, \(P(W \cup O)\).


The formula for the union of two events is \(P(W \cup O) = P(W) + P(O) - P(W \cap O)\).


Total number of people = 8000.

Number of women = 4000.
\(P(W) = \frac{4000}{8000} = \frac{1}{2}\).


Number of people working outside = 3000.
\(P(O) = \frac{3000}{8000} = \frac{3}{8}\).


The event \((W \cap O)\) represents a person who is a woman AND works outside.

Number of women working outside = 30% of 4000 = \(0.30 \times 4000 = 1200\).
\(P(W \cap O) = \frac{1200}{8000} = \frac{12}{80} = \frac{3}{20}\).


Now, we can calculate \(P(W \cup O)\):
\(P(W \cup O) = \frac{1}{2} + \frac{3}{8} - \frac{3}{20}\).


To add/subtract these fractions, we find a common denominator, which is 40.
\(P(W \cup O) = \frac{20}{40} + \frac{15}{40} - \frac{6}{40}\).

\(P(W \cup O) = \frac{20 + 15 - 6}{40} = \frac{29}{40}\).


So, the probability is \(\frac{29}{40}\).
Quick Tip: For "A or B" probability questions, always think of the addition rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). The most common error is forgetting to subtract the probability of the intersection (the overlap).

The problem is to minimize the objective function \(Z = 5x + 3y\).

The shaded feasible region in the diagram corresponds to the inequalities \(x+y \ge 5\), \(x+3y \ge 9\), \(x \ge 0\), and \(y \ge 0\). This is an unbounded region.


To find the minimum value, we first evaluate Z at the corner points of the feasible region.

The corner points are the intersections of the boundary lines.


Point 1: y-intercept of \(x+y=5\). Let \(x=0\), then \(y=5\). The point is (0, 5).

Point 2: Intersection of \(x+y=5\) and \(x+3y=9\).

From \(x+y=5\), we get \(x=5-y\). Substitute into the second equation:
\((5-y) + 3y = 9 \implies 5 + 2y = 9 \implies 2y = 4 \implies y=2\).

Then \(x = 5-2 = 3\). The point is (3, 2).


Point 3: x-intercept of \(x+3y=9\). Let \(y=0\), then \(x=9\). The point is (9, 0).


Now, we evaluate Z at these corner points:

At (0, 5): \(Z = 5(0) + 3(5) = 15\).

At (3, 2): \(Z = 5(3) + 3(2) = 15 + 6 = 21\).

At (9, 0): \(Z = 5(9) + 3(0) = 45\).


The smallest value of Z at a corner point is 15.


Since the feasible region is unbounded, we must check if Z can take a value smaller than 15.

We check the open half-plane \(5x + 3y < 15\).

The boundary line \(5x+3y=15\) passes through (3,0) and (0,5).

Let's check if this half-plane has any point in common with the feasible region.

For example, let's test the corner point (3,2). \(5(3)+3(2) = 21\), which is not less than 15.

Let's test (9,0). \(5(9)+3(0)=45\), not less than 15.

Since the open half-plane \(5x+3y < 15\) does not intersect the feasible region, the minimum value of Z is 15.
Quick Tip: For an unbounded feasible region in a minimization problem, find the minimum value 'm' at the corner points. Then, you must check if the open half-plane defined by \(Z < m\) intersects the feasible region. If it doesn't, 'm' is the minimum value. If it does, there is no minimum value.

A function is bijective if it is both injective (one-to-one) and surjective (onto).


1. Injectivity (One-to-one):

Let \(x_1, x_2 \in A\) such that \(f(x_1) = f(x_2)\).
\(\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}\).


Cross-multiplying gives:
\((x_1-2)(x_2-3) = (x_2-2)(x_1-3)\).
\(x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6\).


Canceling common terms from both sides:
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\).


Rearranging the terms:
\(3x_2 - 2x_2 = 3x_1 - 2x_1\).
\(x_2 = x_1\).

Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function f is injective.


2. Surjectivity (Onto):

Let \(y\) be an arbitrary element in the codomain \(B = R - \{1\}\). We need to find an \(x\) in the domain \(A = R - \{3\}\) such that \(f(x)=y\).
\(y = \frac{x-2}{x-3}\).


We solve for x in terms of y:
\(y(x-3) = x-2\).
\(yx - 3y = x-2\).
\(yx - x = 3y-2\).
\(x(y-1) = 3y-2\).
\(x = \frac{3y-2}{y-1}\).


Since \(y \in B\), \(y \neq 1\), so the denominator \((y-1)\) is never zero. Thus, x is a well-defined real number for every \(y \in B\).

We must also ensure that \(x \in A\), which means \(x \neq 3\).

Suppose \(x=3\). Then \(3 = \frac{3y-2}{y-1} \implies 3y-3 = 3y-2 \implies -3=-2\), which is a contradiction.

This means \(x\) can never be 3.

So, for every \(y \in B\), there exists an \(x = \frac{3y-2}{y-1} \in A\).

Therefore, the function f is surjective.


Since f is both injective and surjective, it is a bijective function.
Quick Tip: To test for surjectivity, set \(y = f(x)\), solve for \(x\) in terms of \(y\), and check two things: 1) Is \(x\) well-defined for all \(y\) in the codomain? 2) Is it possible for this \(x\) to be a value that is excluded from the domain? If \(x\) is always defined and never takes an excluded value, the function is surjective.

The objective function to minimize is \(Z = 18x + 10y\).


The constraints are:

1) \(4x + y \ge 20\)

2) \(2x + 3y \ge 30\)

3) \(x \ge 0, y \ge 0\)


First, we find the corner points of the feasible region by treating the inequalities as equations.

The line \(L_1: 4x+y=20\) passes through \((5,0)\) and \((0,20)\).

The line \(L_2: 2x+3y=30\) passes through \((15,0)\) and \((0,10)\).


The corner points are the intersections of these lines in the first quadrant.

Point A: The y-intercept of \(4x+y=20\). Since the region is \(x \ge 0\), we consider the intersection with the y-axis, but the feasible region is bounded below by \(L_2\) as well. The intersection of \(L_1\) and the y-axis is \((0,20)\).

Point B: The intersection of \(4x+y=20\) and \(2x+3y=30\).

From \(4x+y=20\), we have \(y=20-4x\). Substituting into the second equation:
\(2x + 3(20-4x) = 30\)
\(2x + 60 - 12x = 30\)
\(-10x = -30 \implies x=3\).
\(y = 20 - 4(3) = 20 - 12 = 8\).

So, point B is \((3, 8)\).

Point C: The x-intercept of \(2x+3y=30\), which is \((15, 0)\).


The corner points of the feasible region are A(0, 20), B(3, 8), and C(15, 0).


Now, we evaluate Z at these corner points:

At A(0, 20): \(Z = 18(0) + 10(20) = 200\).

At B(3, 8): \(Z = 18(3) + 10(8) = 54 + 80 = 134\).

At C(15, 0): \(Z = 18(15) + 10(0) = 270\).


The smallest value is 134. Since the feasible region is unbounded, we must check if \(18x+10y < 134\) has any point in common with the feasible region.

The line \(18x+10y=134\) does not intersect the feasible region (other than at the corner point B).

Therefore, the minimum value of Z is 134.
Quick Tip: For an unbounded feasible region in a minimization problem, after finding the lowest value 'm' at a corner point, you must verify that the open half-plane \(Z < m\) does not overlap with the feasible region. If there's no overlap, 'm' is the minimum.

Let the sum of vectors \(\vec{b}\) and \(\vec{c}\) be \(\vec{s}\).
\(\vec{s} = \vec{b} + \vec{c} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\lambda\hat{i} - 2\hat{j} - 3\hat{k})\).
\(\vec{s} = (2+\lambda)\hat{i} + (-4-2)\hat{j} + (5-3)\hat{k} = (2+\lambda)\hat{i} - 6\hat{j} + 2\hat{k}\).


Let \(\hat{s}\) be the unit vector along \(\vec{s}\). Then \(\hat{s} = \frac{\vec{s}}{|\vec{s}|}\).

We are given that the scalar product of \(\vec{a}\) and \(\hat{s}\) is 1.
\(\vec{a} \cdot \hat{s} = 1 \implies \frac{\vec{a} \cdot \vec{s}}{|\vec{s}|} = 1 \implies \vec{a} \cdot \vec{s} = |\vec{s}|\).


First, calculate the dot product \(\vec{a} \cdot \vec{s}\):
\(\vec{a} \cdot \vec{s} = (\hat{i} - \hat{j} + 2\hat{k}) \cdot ((2+\lambda)\hat{i} - 6\hat{j} + 2\hat{k})\).
\(\vec{a} \cdot \vec{s} = (1)(2+\lambda) + (-1)(-6) + (2)(2) = 2+\lambda+6+4 = \lambda+12\).


Next, calculate the magnitude \(|\vec{s}|\):
\(|\vec{s}| = \sqrt{(2+\lambda)^2 + (-6)^2 + 2^2} = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40}\).


Now, set \(\vec{a} \cdot \vec{s} = |\vec{s}|\):
\(\lambda+12 = \sqrt{(2+\lambda)^2 + 40}\).


Square both sides:
\((\lambda+12)^2 = (2+\lambda)^2 + 40\).
\(\lambda^2 + 24\lambda + 144 = \lambda^2 + 4\lambda + 4 + 40\).
\(24\lambda + 144 = 4\lambda + 44\).
\(20\lambda = -100\).
\(\lambda = -5\).
Quick Tip: The condition that the scalar product of a vector \(\vec{a}\) with a unit vector \(\hat{b}\) is 'k' can be written as \(\vec{a} \cdot \hat{b} = k\), or more usefully as \(\vec{a} \cdot \vec{b} = k|\vec{b}|\). This avoids dealing with square roots in the denominator initially.

The given lines are in the form \(\vec{r} = \vec{a_1} + \lambda\vec{b_1}\) and \(\vec{r} = \vec{a_2} + \mu\vec{b_2}\).

Here, \(\vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{b_1} = \hat{i} - 2\hat{j} + 3\hat{k}\).

And \(\vec{a_2} = \hat{i} + 4\hat{k}\) and \(\vec{b_2} = 3\hat{i} - 6\hat{j} + 9\hat{k}\).


First, we check if the lines are parallel by comparing their direction vectors, \(\vec{b_1}\) and \(\vec{b_2}\).
\(\vec{b_2} = 3\hat{i} - 6\hat{j} + 9\hat{k} = 3(\hat{i} - 2\hat{j} + 3\hat{k}) = 3\vec{b_1}\).

Since \(\vec{b_2}\) is a scalar multiple of \(\vec{b_1}\), the lines are parallel.


The shortest distance between two parallel lines is given by the formula:
\(d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}|}{|\vec{b_1}|}\).


First, find \(\vec{a_2} - \vec{a_1}\):
\(\vec{a_2} - \vec{a_1} = (\hat{i} + 4\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) = (1-2)\hat{i} - (-1)\hat{j} + (4-3)\hat{k} = -\hat{i} + \hat{j} + \hat{k}\).


Next, find the cross product \((\vec{a_2} - \vec{a_1}) \times \vec{b_1}\):
\((\vec{a_2} - \vec{a_1}) \times \vec{b_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 1
1 & -2 & 3 \end{vmatrix}\).
\(= \hat{i}(3 - (-2)) - \hat{j}(-3 - 1) + \hat{k}(2 - 1) = 5\hat{i} + 4\hat{j} + \hat{k}\).


Find the magnitude of this cross product:
\(|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}| = \sqrt{5^2 + 4^2 + 1^2} = \sqrt{25 + 16 + 1} = \sqrt{42}\).


Find the magnitude of \(\vec{b_1}\):
\(|\vec{b_1}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}\).


Finally, calculate the distance d:
\(d = \frac{\sqrt{42}}{\sqrt{14}} = \sqrt{\frac{42}{14}} = \sqrt{3}\) units.
Quick Tip: Always start by comparing the direction vectors (\(\vec{b_1}, \vec{b_2}\)) to see if the lines are parallel. The formula for the shortest distance is different for parallel lines and skew lines. For parallel lines, use \(d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}|}{|\vec{b_1}|}\).

Let \(y = \log(x^x + \csc^2 x)\).


Using the chain rule, \(\frac{dy}{dx} = \frac{1}{x^x + \csc^2 x} \cdot \frac{d}{dx}(x^x + \csc^2 x)\).


We need to find the derivatives of \(x^x\) and \(\csc^2 x\) separately.


Let \(u = x^x\). To differentiate this, we use logarithmic differentiation.
\(\ln u = \ln(x^x) = x \ln x\).

Differentiating both sides with respect to x:
\(\frac{1}{u}\frac{du}{dx} = (1)(\ln x) + (x)(\frac{1}{x}) = \ln x + 1\).
\(\frac{du}{dx} = u(\ln x + 1) = x^x(1 + \ln x)\).


Let \(v = \csc^2 x = (\csc x)^2\). We use the chain rule here.
\(\frac{dv}{dx} = 2(\csc x) \cdot \frac{d}{dx}(\csc x)\).
\(\frac{d}{dx}(\csc x) = -\csc x \cot x\).

So, \(\frac{dv}{dx} = 2\csc x (-\csc x \cot x) = -2 \csc^2 x \cot x\).


Now, combine the derivatives:
\(\frac{d}{dx}(x^x + \csc^2 x) = \frac{du}{dx} + \frac{dv}{dx} = x^x(1 + \ln x) - 2 \csc^2 x \cot x\).


Finally, substitute this back into the first equation:
\(\frac{dy}{dx} = \frac{x^x(1 + \ln x) - 2 \csc^2 x \cot x}{x^x + \csc^2 x}\).
Quick Tip: For functions of the form \(f(x)^{g(x)}\), logarithmic differentiation is the standard method. Take the natural log of both sides, use log properties to bring the exponent down, and then differentiate implicitly.

Let the fixed perimeter of the rectangle be P.

Let the length and breadth of the rectangle be x and y, respectively.


Perimeter, \(P = 2(x+y)\). This is a constant.

From this, we can express y in terms of x: \(y = \frac{P}{2} - x\).


The area of the rectangle is \(A = xy\).

Substituting the expression for y, we get the area A as a function of x:
\(A(x) = x(\frac{P}{2} - x) = \frac{P}{2}x - x^2\).


To find the maximum area, we use the first derivative test. We differentiate A with respect to x:
\(\frac{dA}{dx} = \frac{P}{2} - 2x\).


Set the first derivative to zero to find critical points:
\(\frac{dA}{dx} = 0 \implies \frac{P}{2} - 2x = 0 \implies 2x = \frac{P}{2} \implies x = \frac{P}{4}\).


Now, we find the corresponding value of y:
\(y = \frac{P}{2} - x = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}\).


Since \(x = y = \frac{P}{4}\), the rectangle is a square.


To confirm that this corresponds to a maximum area, we use the second derivative test.
\(\frac{d^2A}{dx^2} = \frac{d}{dx}(\frac{P}{2} - 2x) = -2\).


Since \(\frac{d^2A}{dx^2} = -2 < 0\), the area is maximum when \(x = \frac{P}{4}\).

Thus, for a fixed perimeter, the area is greatest when the rectangle is a square.
Quick Tip: In optimization problems, the process is to express the quantity to be optimized (e.g., area) as a function of a single variable, find the critical points by setting the first derivative to zero, and then use the second derivative test to confirm if it's a maximum or minimum.

To show the function is bijective, we must prove it is both injective (one-one) and surjective (onto).


Injectivity (One-one):

Let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\).
\(4x_1^3 - 5 = 4x_2^3 - 5\).
\(4x_1^3 = 4x_2^3\).
\(x_1^3 = x_2^3\).

Taking the cube root on both sides, we get \(x_1 = x_2\).

Since \(f(x_1) = f(x_2) \implies x_1 = x_2\), the function is one-one.


Surjectivity (Onto):

Let \(y\) be an arbitrary element in the codomain R. We need to find if there exists an \(x\) in the domain R such that \(f(x) = y\).
\(y = 4x^3 - 5\).

We solve for x:
\(y+5 = 4x^3\).
\(x^3 = \frac{y+5}{4}\).
\(x = \sqrt[3]{\frac{y+5}{4}}\).

For any real number y, \(\frac{y+5}{4}\) is a real number, and its real cube root is also a unique real number.

So, for every \(y\) in the codomain, there exists a pre-image \(x\) in the domain.

Therefore, the function is onto.


Since the function is both one-one and onto, it is bijective.
Quick Tip: An alternative way to prove injectivity for a differentiable function is to show its derivative is either always non-negative or always non-positive. Here, \(f'(x) = 12x^2 \ge 0\), which means the function is always non-decreasing, making it injective.

To be an equivalence relation, R must be reflexive, symmetric, and transitive.


Reflexivity:

Let \(x \in N\). We need to check if \((x,x) \in R\).

The product is \(x \cdot x = x^2\).

Since x is a natural number, \(x^2\) is the square of a natural number.

Thus, \((x,x) \in R\) for all \(x \in N\). So, R is reflexive.


Symmetry:

Let \((x,y) \in R\). This means \(xy = k^2\) for some natural number k.

By the commutative property of multiplication, \(yx = xy = k^2\).

Since \(yx\) is the square of a natural number, \((y,x) \in R\).

Thus, if \((x,y) \in R\), then \((y,x) \in R\). So, R is symmetric.


Transitivity:

Let \((x,y) \in R\) and \((y,z) \in R\). We need to check if \((x,z) \in R\).
\((x,y) \in R \implies xy = k^2\) for some \(k \in N\).
\((y,z) \in R \implies yz = m^2\) for some \(m \in N\).

Multiplying these two equations:

\((xy)(yz) = k^2m^2 \implies xz \cdot y^2 = (km)^2\).
\(xz = \frac{(km)^2}{y^2} = (\frac{km}{y})^2\).


Since \(x, y, z, k, m\) are natural numbers, \(\frac{km}{y}\) is a rational number. For \(xz\) to be the square of a natural number, \(\frac{km}{y}\) must be a natural number.


From \(xy=k^2\) and \(yz=m^2\), y must divide both \(k^2\) and \(m^2\). Let the prime factorization of any number \(n\) be written as \(s \cdot p\) where \(s\) is the square-free part. For \(xy\) to be a perfect square, \(x\) and \(y\) must have the same square-free part.


If \(x\) and \(y\) have the same square-free part, and \(y\) and \(z\) have the same square-free part, then \(x\) and \(z\) must also have the same square-free part. Therefore, their product \(xz\) will be a perfect square.
Thus, if \((x,y) \in R\) and \((y,z) \in R\), then \((x,z) \in R\). So, R is transitive.


Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Quick Tip: A useful way to think about this relation is that \((x,y) \in R\) if the product of prime factors with odd powers is the same for both x and y. This property is reflexive, symmetric, and transitive.

The system of linear equations is:
\(2x + 5y = 1\)
\(3x + 2y = 7\)


This system can be written in the matrix form \(AX = B\), where:
\(A = \begin{bmatrix} 2 & 5
3 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x
y \end{bmatrix}\), and \(B = \begin{bmatrix} 1
7 \end{bmatrix}\).


The solution is given by \(X = A^{-1}B\).

First, we find the determinant of A:
\(|A| = (2)(2) - (5)(3) = 4 - 15 = -11\).

Since \(|A| \neq 0\), the inverse exists.


Next, we find the adjoint of A:

adj(A) = \(\begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix}\).


Now, we find the inverse of A:
\(A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-11} \begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix}\).


Finally, we calculate X:
\(X = A^{-1}B = \frac{1}{-11} \begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix} \begin{bmatrix} 1
7 \end{bmatrix}\).
\(X = -\frac{1}{11} \begin{bmatrix} (2)(1) + (-5)(7)
(-3)(1) + (2)(7) \end{bmatrix}\).
\(X = -\frac{1}{11} \begin{bmatrix} 2 - 35
-3 + 14 \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} -33
11 \end{bmatrix}\).
\(X = \begin{bmatrix} \frac{-33}{-11}
\frac{11}{-11} \end{bmatrix} = \begin{bmatrix} 3
-1 \end{bmatrix}\).


So, \(x=3\) and \(y=-1\).

The common point is \((3, -1)\).
Quick Tip: For solving a \(2 \times 2\) system using the matrix method, remember the quick formula for the inverse: for a matrix \(A = \begin{bmatrix} a & b
c & d \end{bmatrix}\), the inverse is \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b
-c & a \end{bmatrix}\). This avoids calculating cofactors explicitly.

Let the sales of books be represented by a \(2 \times 3\) matrix S.
\(S = \begin{bmatrix} 50 & 60 & 35
40 & 45 & 50 \end{bmatrix} \begin{matrix} Day I
Day II \end{matrix}\).


Let the selling prices be represented by a \(3 \times 1\) column matrix P.
\(P = \begin{bmatrix} 150
175
180 \end{bmatrix} \begin{matrix} Chemistry
Physics
Maths \end{matrix}\).


The total sales for each day can be found by the matrix product SP.
\(SP = \begin{bmatrix} 50 & 60 & 35
40 & 45 & 50 \end{bmatrix} \begin{bmatrix} 150
175
180 \end{bmatrix}\).

\(SP = \begin{bmatrix} (50)(150) + (60)(175) + (35)(180)
(40)(150) + (45)(175) + (50)(180) \end{bmatrix}\).

\(SP = \begin{bmatrix} 7500 + 10500 + 6300
6000 + 7875 + 9000 \end{bmatrix} = \begin{bmatrix} 24300
22875 \end{bmatrix}\).

This matrix shows the sales for Day I (Rs 24,300) and Day II (Rs 22,875).


The total sale over the two days is the sum of the elements of the resulting matrix.

Total Sale = \(24300 + 22875 = Rs 47,175\).


The total cost price of all the books is given as Rs 35,000.

Profit = Total Sale - Total Cost Price.

Profit = \(47,175 - 35,000 = Rs 12,175\).
Quick Tip: When setting up matrix multiplication for real-world problems, ensure the dimensions are compatible. If you have a \(m \times n\) matrix for quantities and want to multiply by prices, the price matrix must be \(n \times p\). The resulting \(m \times p\) matrix will have a meaningful interpretation.

We use the method of partial fraction decomposition.


Let the integrand be expressed as:
\(\frac{3x+1}{(x-2)^2(x+2)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2}\).


Multiplying both sides by the denominator \((x-2)^2(x+2)\), we get:
\(3x+1 = A(x-2)(x+2) + B(x+2) + C(x-2)^2\).


To find the constants A, B, and C, we substitute strategic values for x.

Let \(x=2\):
\(3(2)+1 = A(0) + B(2+2) + C(0) \implies 7 = 4B \implies B = \frac{7}{4}\).


Let \(x=-2\):
\(3(-2)+1 = A(0) + B(0) + C(-2-2)^2 \implies -5 = 16C \implies C = -\frac{5}{16}\).


To find A, we can equate the coefficients of \(x^2\) on both sides.

The coefficient of \(x^2\) on the LHS is 0. On the RHS, it is \(A+C\).
\(0 = A+C \implies A = -C \implies A = \frac{5}{16}\).


Now, we can integrate the partial fractions:
\(\int (\frac{5/16}{x-2} + \frac{7/4}{(x-2)^2} - \frac{5/16}{x+2}) dx\).

\(= \frac{5}{16}\int \frac{1}{x-2}dx + \frac{7}{4}\int (x-2)^{-2}dx - \frac{5}{16}\int \frac{1}{x+2}dx\).

\(= \frac{5}{16}\ln|x-2| + \frac{7}{4}\frac{(x-2)^{-1}}{-1} - \frac{5}{16}\ln|x+2| + C\).

\(= \frac{5}{16}(\ln|x-2| - \ln|x+2|) - \frac{7}{4(x-2)} + C\).

\(= \frac{5}{16}\ln|\frac{x-2}{x+2}| - \frac{7}{4(x-2)} + C\).
Quick Tip: For partial fractions with repeated linear factors like \((x-a)^2\), the decomposition must include terms for each power: \(\frac{A}{x-a} + \frac{B}{(x-a)^2}\). The substitution method is fastest for finding constants corresponding to the highest powers and non-repeated factors.

Let \(I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x} dx\) --- (1).


Using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\):
\(I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} dx\).

\(I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\cos x + \sin x} dx\) --- (2).


Adding equations (1) and (2):
\(2I = \int_{0}^{\pi/2} \frac{x + (\frac{\pi}{2} - x)}{\sin x + \cos x} dx = \int_{0}^{\pi/2} \frac{\frac{\pi}{2}}{\sin x + \cos x} dx\).

\(2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} dx\).


To evaluate the integral, we write the denominator in the form \(R\sin(x+\alpha)\).
\(\sin x + \cos x = \sqrt{2}(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x) = \sqrt{2}(\cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x) = \sqrt{2}\sin(x+\frac{\pi}{4})\).

\(2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})} dx = \frac{\pi}{2\sqrt{2}} \int_{0}^{\pi/2} \csc(x+\frac{\pi}{4}) dx\).

\(2I = \frac{\pi}{2\sqrt{2}} [-\ln|\csc(x+\frac{\pi}{4}) + \cot(x+\frac{\pi}{4})|]_{0}^{\pi/2}\).

\(2I = -\frac{\pi}{2\sqrt{2}} [(\ln|\csc\frac{3\pi}{4} + \cot\frac{3\pi}{4}|) - (\ln|\csc\frac{\pi}{4} + \cot\frac{\pi}{4}|)]\).

\(2I = -\frac{\pi}{2\sqrt{2}} [\ln|\sqrt{2} - 1| - \ln|\sqrt{2} + 1|] = -\frac{\pi}{2\sqrt{2}} \ln(\frac{\sqrt{2}-1}{\sqrt{2}+1})\).

\(2I = -\frac{\pi}{2\sqrt{2}} \ln(\frac{(\sqrt{2}-1)^2}{1}) = -\frac{\pi}{2\sqrt{2}} (2\ln(\sqrt{2}-1)) = \frac{\pi}{\sqrt{2}}\ln(\frac{1}{\sqrt{2}-1}) = \frac{\pi}{\sqrt{2}}\ln(\sqrt{2}+1)\).

\(I = \frac{\pi}{2\sqrt{2}}\ln(\sqrt{2}+1)\).
Quick Tip: When you see an integral of the form \(\int_0^a \frac{x \cdot g(x)}{h(x)} dx\) where \(h(a-x)=h(x)\), applying the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\) and adding the two integrals is a very effective strategy to eliminate the 'x' term from the numerator.

First, we convert the equation of the line to its standard form.
\(\frac{2(x+2)}{6} = \frac{y+1}{2} = \frac{-2(z-3)}{-4}\).
\(\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}\).


Let this common ratio be \(\lambda\). Any general point Q on this line can be represented as:
\(Q(3\lambda-2, 2\lambda-1, 2\lambda+3)\).


We are given the point P(1, 2, 3) and the distance \(PQ = 3\sqrt{2}\).

Using the distance formula, \((PQ)^2 = (3\sqrt{2})^2 = 18\).
\(((3\lambda-2)-1)^2 + ((2\lambda-1)-2)^2 + ((2\lambda+3)-3)^2 = 18\).
\((3\lambda-3)^2 + (2\lambda-3)^2 + (2\lambda)^2 = 18\).
\(9(\lambda-1)^2 + (4\lambda^2 - 12\lambda + 9) + 4\lambda^2 = 18\).
\(9(\lambda^2 - 2\lambda + 1) + 4\lambda^2 - 12\lambda + 9 + 4\lambda^2 = 18\).
\(9\lambda^2 - 18\lambda + 9 + 8\lambda^2 - 12\lambda + 9 = 18\).
\(17\lambda^2 - 30\lambda + 18 = 18\).
\(17\lambda^2 - 30\lambda = 0\).
\(\lambda(17\lambda - 30) = 0\).


This gives two possible values for \(\lambda\): \(\lambda = 0\) or \(\lambda = \frac{30}{17}\).


Case 1: If \(\lambda = 0\).

The point Q is \((3(0)-2, 2(0)-1, 2(0)+3)\), which is \((-2, -1, 3)\).


Case 2: If \(\lambda = \frac{30}{17}\).

The point Q is \((3(\frac{30}{17})-2, 2(\frac{30}{17})-1, 2(\frac{30}{17})+3)\), which is \((\frac{56}{17}, \frac{43}{17}, \frac{111}{17})\).


Both points are valid. We can state one of them, for instance, Q(-2, -1, 3).
Quick Tip: Always start by converting the equation of a line into the standard symmetric form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\). This makes it easy to write the coordinates of a general point on the line using a parameter \(\lambda\).

Let the given point be P(-1, 5, 2).

First, write the line equation in standard form:
\(\frac{2(x-2)}{2} = \frac{y}{2} = \frac{-(z-2)}{3} \implies \frac{x-2}{1} = \frac{y}{2} = \frac{z-2}{-3}\).


Let M be the foot of the perpendicular from P to the line. Any point M on the line can be written as \(M(\lambda+2, 2\lambda, -3\lambda+2)\).

The direction ratios of the line segment PM are \((\lambda+2 - (-1), 2\lambda - 5, -3\lambda+2 - 2)\), which simplifies to \((\lambda+3, 2\lambda-5, -3\lambda)\).


The direction ratios of the given line are \((1, 2, -3)\).

Since PM is perpendicular to the line, the dot product of their direction ratios is zero.
\(1(\lambda+3) + 2(2\lambda-5) + (-3)(-3\lambda) = 0\).
\(\lambda+3 + 4\lambda-10 + 9\lambda = 0\).
\(14\lambda - 7 = 0 \implies \lambda = \frac{1}{2}\).


The coordinates of the foot of the perpendicular M are:
\(M(\frac{1}{2}+2, 2(\frac{1}{2}), -3(\frac{1}{2})+2) = (\frac{5}{2}, 1, \frac{1}{2})\).


Let the image of P be P'(x', y', z'). M is the midpoint of PP'.
\(\frac{x' + (-1)}{2} = \frac{5}{2} \implies x' - 1 = 5 \implies x' = 6\).
\(\frac{y' + 5}{2} = 1 \implies y' + 5 = 2 \implies y' = -3\).
\(\frac{z' + 2}{2} = \frac{1}{2} \implies z' + 2 = 1 \implies z' = -1\).

The image point is P'(6, -3, -1).


The length of the line segment joining P and P' is the distance between them.
\(PP' = \sqrt{(6 - (-1))^2 + (-3 - 5)^2 + (-1 - 2)^2}\).
\(PP' = \sqrt{7^2 + (-8)^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122}\) units.
Quick Tip: Finding the image of a point in a line involves three steps: 1. Find the general coordinates of the foot of the perpendicular (M) on the line. 2. Use the dot product of direction ratios of PM and the line to find the parameter. 3. Use the midpoint formula, as M is the midpoint of the point and its image.

The given differential equation is \((x - \sin y) dy + (\tan y) dx = 0\).


Rearranging the terms to express it as a linear differential equation in x:
\((\tan y) \frac{dx}{dy} + x - \sin y = 0\).
\(\frac{dx}{dy} + \frac{x}{\tan y} - \frac{\sin y}{\tan y} = 0\).
\(\frac{dx}{dy} + (\cot y)x = \cos y\).


This is a linear differential equation of the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = \cot y\) and \(Q(y) = \cos y\).


The integrating factor (I.F.) is given by \(e^{\int P(y) dy}\).

I.F. = \(e^{\int \cot y dy} = e^{\ln|\sin y|} = \sin y\).


The general solution is given by \(x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C\).
\(x \sin y = \int \cos y \sin y dy + C\).


To evaluate the integral, we can write it as \(\frac{1}{2}\int 2\sin y \cos y dy = \frac{1}{2}\int \sin(2y) dy\).
\(\int \cos y \sin y dy = \frac{1}{2} (-\frac{\cos(2y)}{2}) = -\frac{\cos(2y)}{4}\).


So, the general solution is \(x \sin y = -\frac{\cos(2y)}{4} + C\).


We are given the initial condition \(y(0) = 0\), which means \(y=0\) when \(x=0\).

Substitute these values to find C:
\(0 \cdot \sin(0) = -\frac{\cos(0)}{4} + C\).
\(0 = -\frac{1}{4} + C \implies C = \frac{1}{4}\).


The particular solution is \(x \sin y = -\frac{\cos(2y)}{4} + \frac{1}{4}\).

Multiplying by 4 gives \(4x \sin y = 1 - \cos(2y)\).

Using the identity \(1 - \cos(2y) = 2\sin^2 y\):
\(4x \sin y = 2\sin^2 y\).

This equation gives two possibilities: \(\sin y = 0\) or \(4x = 2\sin y\).

The solution is \(\sin y = 2x\). (The solution \(\sin y = 0\) is a singular solution).
Quick Tip: When a differential equation is not linear in the form \(\frac{dy}{dx}\), check if it becomes linear by considering x as the dependent variable and y as the independent variable, i.e., look for the form \(\frac{dx}{dy} + P(y)x = Q(y)\).

The equation of the circular table top centered at the origin is \(x^2 + y^2 = 8^2 = 64\).

The boundaries of the region in the first quadrant are:

1. The x-axis: \(y=0\).

2. The scratch: a line through the origin with an angle of \(\frac{\pi}{4}\), so its equation is \(y = x\).

3. The circle: \(y = \sqrt{64 - x^2}\).


To find the area using integration, we can split the region into two parts at the point where the line \(y=x\) intersects the circle.

Intersection point: \(x^2 + x^2 = 64 \implies 2x^2 = 64 \implies x^2 = 32 \implies x = 4\sqrt{2}\).


The total area A is the sum of two integrals:
\(A = \int_{0}^{4\sqrt{2}} x \,dx + \int_{4\sqrt{2}}^{8} \sqrt{64-x^2} \,dx\).


Part 1: Area under the line \(y=x\).
\(\int_{0}^{4\sqrt{2}} x \,dx = [\frac{x^2}{2}]_{0}^{4\sqrt{2}} = \frac{(4\sqrt{2})^2}{2} - 0 = \frac{32}{2} = 16\).


Part 2: Area under the circle. We use the formula \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})\).
\(\int_{4\sqrt{2}}^{8} \sqrt{64-x^2} \,dx = [\frac{x}{2}\sqrt{64-x^2} + \frac{64}{2}\sin^{-1}(\frac{x}{8})]_{4\sqrt{2}}^{8}\).


Evaluating at the upper limit (\(x=8\)):
\(\frac{8}{2}\sqrt{64-64} + 32\sin^{-1}(\frac{8}{8}) = 0 + 32(\frac{\pi}{2}) = 16\pi\).


Evaluating at the lower limit (\(x=4\sqrt{2}\)):
\(\frac{4\sqrt{2}}{2}\sqrt{64-32} + 32\sin^{-1}(\frac{4\sqrt{2}}{8}) = 2\sqrt{2}\sqrt{32} + 32\sin^{-1}(\frac{1}{\sqrt{2}})\).
\(= 2\sqrt{64} + 32(\frac{\pi}{4}) = 16 + 8\pi\).


The value of the second integral is \((16\pi) - (16 + 8\pi) = 8\pi - 16\).


Total Area A = (Part 1) + (Part 2) = \(16 + (8\pi - 16) = 8\pi\) cm\(^2\).
Quick Tip: While integration is required here, you can verify your answer using the geometric formula for the area of a sector: \(A = \frac{1}{2}r^2\theta\). With \(r=8\) and \(\theta=\pi/4\), the area is \(\frac{1}{2}(8^2)(\frac{\pi}{4}) = 8\pi\). This confirms the result.

The given differential equation is \(\frac{dV}{dt} = kS\).


The order of a differential equation is the order of the highest derivative present in the equation.

In this equation, the highest derivative is \(\frac{dV}{dt}\), which is a first-order derivative.

Therefore, the order of the differential equation is 1.


The degree of a differential equation is the highest power of the highest-order derivative, after the equation has been cleared of radicals and fractions in its derivatives.

The highest derivative \(\frac{dV}{dt}\) has a power of 1.

Therefore, the degree of the differential equation is 1.
Quick Tip: The order is determined by the highest derivative (e.g., \(\frac{d^2y}{dx^2}\) is order 2), while the degree is the power of that highest derivative.

The given differential equation is \(\frac{dr}{dt} = \frac{2}{3}k\).


This is a separable differential equation. We can write it as \(dr = (\frac{2}{3}k) dt\).


Integrating both sides:
\(\int dr = \int \frac{2}{3}k \, dt\).

\(r = (\frac{2}{3}k)t + C\), where C is the constant of integration.


We are given the initial condition \(r(0) = 5\) mm. This means when \(t=0\), \(r=5\).


Substituting these values into the general solution to find C:
\(5 = (\frac{2}{3}k)(0) + C\).

\(C = 5\).


Therefore, the particular solution to the differential equation is:
\(r(t) = \frac{2}{3}kt + 5\).
Quick Tip: When solving a simple differential equation like \(\frac{dy}{dx} = K\) (where K is a constant), the solution is always a linear function \(y = Kx + C\). Use the initial condition to find the value of the integration constant C.

From the previous part, we have the solution \(r(t) = \frac{2}{3}kt + 5\).


We are given that \(r=3\) when \(t=1\). Substitute these values into the equation:
\(3 = \frac{2}{3}k(1) + 5\).

\(3 - 5 = \frac{2}{3}k\).

\(-2 = \frac{2}{3}k\).

\(k = -2 \times \frac{3}{2} = -3\).


Now we have the specific equation for the radius as a function of time:
\(r(t) = \frac{2}{3}(-3)t + 5 \implies r(t) = -2t + 5\).


We need to find the time 't' when the radius becomes 0, i.e., \(r=0\).
\(0 = -2t + 5\).

\(2t = 5\).

\(t = \frac{5}{2} = 2.5\) hours.
Quick Tip: After finding the general solution to a differential equation, use the given data points (like \(r=3\) at \(t=1\)) to solve for the unknown parameters (like \(k\)). Then use the fully determined equation to answer further questions.

We start with the solution from part (ii): \(r(t) = \frac{2}{3}kt + 5\).


We are given that \(r=1\) when \(t=1\). Substitute these values:
\(1 = \frac{2}{3}k(1) + 5\).

\(1 - 5 = \frac{2}{3}k\).

\(-4 = \frac{2}{3}k\).

\(k = -4 \times \frac{3}{2} = -6\).


Now the specific equation for the radius is:
\(r(t) = \frac{2}{3}(-6)t + 5 \implies r(t) = -4t + 5\).


We need to find the time 't' when the radius becomes 0, i.e., \(r=0\).
\(0 = -4t + 5\).

\(4t = 5\).

\(t = \frac{5}{4} = 1.25\) hours.
Quick Tip: This problem follows the same structure as the previous one. A different data point leads to a different value for the constant 'k', which in turn changes the final prediction. Always be careful with substitutions.

Let \(A_1\), \(A_2\), and \(A_3\) be the events that a person has good, average, and poor health, respectively.

Let D be the event that a person has contracted the disease.


From the problem statement, we have the following probabilities:
\(P(A_1) = \frac{700}{1000} = 0.7\).
\(P(A_2) = \frac{200}{1000} = 0.2\).
\(P(A_3) = \frac{100}{1000} = 0.1\).


The conditional probabilities of contracting the disease are:
\(P(D|A_1) = 25% = 0.25\).
\(P(D|A_2) = 35% = 0.35\).
\(P(D|A_3) = 50% = 0.50\).


Using the Law of Total Probability, the probability of contracting the disease is:
\(P(D) = P(A_1)P(D|A_1) + P(A_2)P(D|A_2) + P(A_3)P(D|A_3)\).

\(P(D) = (0.7)(0.25) + (0.2)(0.35) + (0.1)(0.50)\).

\(P(D) = 0.175 + 0.070 + 0.050\).

\(P(D) = 0.295\).
Quick Tip: The Law of Total Probability is used to find the probability of an event (like contracting a disease) by summing the probabilities of its occurrences across a set of mutually exclusive and exhaustive partitions (like health categories).

We need to find the probability \(P(A_2 | D')\), where \(D'\) is the event that a person has not contracted the disease.


We use Bayes' Theorem: \(P(A_2 | D') = \frac{P(D' | A_2)P(A_2)}{P(D')}\).


First, we find the required conditional probabilities of not contracting the disease:
\(P(D' | A_2) = 1 - P(D | A_2) = 1 - 0.35 = 0.65\).


We also need the prior probability \(P(A_2) = 0.2\).


Next, we find the total probability of not contracting the disease, \(P(D')\).
\(P(D') = 1 - P(D)\). From the previous part, \(P(D)=0.295\).
\(P(D') = 1 - 0.295 = 0.705\).


Now, substitute these values into Bayes' Theorem:
\(P(A_2 | D') = \frac{(0.65)(0.2)}{0.705}\).

\(P(A_2 | D') = \frac{0.130}{0.705}\).


To simplify the fraction, multiply the numerator and denominator by 1000:
\(P(A_2 | D') = \frac{130}{705} = \frac{26 \times 5}{141 \times 5} = \frac{26}{141}\).
Quick Tip: Bayes' Theorem is used to update the probability of a cause (being in health category A2) given a new piece of evidence (not contracting the disease). The formula is: \(P(Cause|Effect) = \frac{P(Effect|Cause)P(Cause)}{P(Effect)}\).

The problem states that friends A and B will meet C at his destination.

This means A travels from A to C, and B travels from B to C.


The movement from A to C is represented by the vector \(\vec{AC}\).

The movement from B to C is represented by the vector \(\vec{BC}\).


To complete the figure, we draw directed line segments (arrows) starting from point A and ending at point C, and starting from point B and ending at point C.

The completed figure shows the initial paths from O (\(\vec{a}\) and \(\vec{b}\)) and the subsequent paths to the meeting point C (\(\vec{AC}\) and \(\vec{BC}\)).
Quick Tip: A vector representing movement from point P to point Q is written as \(\vec{PQ}\). Geometrically, it's an arrow with its tail at P and its head at Q.

We are given the position vectors of points A, B, and C with respect to the origin O:
\(\vec{OA} = \vec{a}\)
\(\vec{OB} = \vec{b}\)
\(\vec{OC} = 5\vec{a} - 2\vec{b}\)


The vector from point A to point C is given by the difference of their position vectors:
\(\vec{AC} = \vec{OC} - \vec{OA}\).

\(\vec{AC} = (5\vec{a} - 2\vec{b}) - (\vec{a})\).
\(\vec{AC} = (5-1)\vec{a} - 2\vec{b} = 4\vec{a} - 2\vec{b}\).


Similarly, the vector from point B to point C is:
\(\vec{BC} = \vec{OC} - \vec{OB}\).

\(\vec{BC} = (5\vec{a} - 2\vec{b}) - (\vec{b})\).
\(\vec{BC} = 5\vec{a} + (-2-1)\vec{b} = 5\vec{a} - 3\vec{b}\).
Quick Tip: The vector from a point P to a point Q can always be found by subtracting the position vector of the starting point from the position vector of the ending point: \(\vec{PQ} = \vec{OQ} - \vec{OP}\).

We are given:

Distance O to A = \(|\vec{OA}| = |\vec{a}| = 1\) km.

Distance O to B = \(|\vec{OB}| = |\vec{b}| = 2\) km.
\(\vec{a} \cdot \vec{b} = 1\).


Let \(\theta\) be the angle between vectors \(\vec{a}\) and \(\vec{b}\). The formula for the dot product is:
\(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\).


Substituting the given values:
\(1 = (1)(2) \cos\theta\).
\(1 = 2 \cos\theta \implies \cos\theta = \frac{1}{2}\).

The angle is \(\theta = \arccos(\frac{1}{2}) = \frac{\pi}{3}\) radians or \(60^\circ\).


Now, we find the magnitude of the cross product, \(|\vec{a} \times \vec{b}|\).

The formula is \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta\).


We first need to find \(\sin\theta\). Since \(\cos\theta = 1/2\), and \(\theta\) is an angle between vectors, we can assume \(0 \le \theta \le \pi\), so \(\sin\theta \ge 0\).
\(\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).


Now, substitute the values into the cross product magnitude formula:
\(|\vec{a} \times \vec{b}| = (1)(2)(\frac{\sqrt{3}}{2}) = \sqrt{3}\).
Quick Tip: Remember the geometric interpretations of the dot product and cross product. The dot product relates to the cosine of the angle between vectors, while the magnitude of the cross product relates to the sine of the angle and represents the area of the parallelogram formed by the vectors.

First, we find the vectors \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\).
\(\vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + (\hat{j} - \hat{k}) = 2\hat{i} + 0\hat{j} + 3\hat{k}\).
\(\vec{a} - \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{j} - \hat{k}) = 2\hat{i} - 2\hat{j} + 5\hat{k}\).


A vector perpendicular to both \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is given by their cross product. Let's call this vector \(\vec{v}\).
\(\vec{v} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\).

\(\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 0 & 3
2 & -2 & 5 \end{vmatrix}\).

\(\vec{v} = \hat{i}((0)(5) - (3)(-2)) - \hat{j}((2)(5) - (3)(2)) + \hat{k}((2)(-2) - (0)(2))\).

\(\vec{v} = \hat{i}(0 - (-6)) - \hat{j}(10 - 6) + \hat{k}(-4 - 0)\).
\(\vec{v} = 6\hat{i} - 4\hat{j} - 4\hat{k}\).


To find the unit vector, we first find the magnitude of \(\vec{v}\).
\(|\vec{v}| = \sqrt{6^2 + (-4)^2 + (-4)^2} = \sqrt{36 + 16 + 16} = \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}\).


The unit vector \(\hat{v}\) is \(\frac{\vec{v}}{|\vec{v}|}\).
\(\hat{v} = \frac{6\hat{i} - 4\hat{j} - 4\hat{k}}{2\sqrt{17}} = \frac{3\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{17}}\).
Quick Tip: A vector perpendicular to two given vectors \(\vec{p}\) and \(\vec{q}\) is always in the direction of their cross product, \(\vec{p} \times \vec{q}\). A unit vector is found by dividing the vector by its own magnitude. Note that \((\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = -2(\vec{a} \times \vec{b})\), which provides a shortcut if you calculate \(\vec{a} \times \vec{b}\) first.